Question

If $n$ is an integer with $0\le n\le 11,$ then the minimum value of $n!(11-n)!$ is attained when a value of $n$ is equal to?

• A. $11$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is B

$5$

Explanation for the correct option:

Let $y=n!(11-n)!$

Consider $x={}^{11}C_n=\frac{11!}{n!(11-n)!}$

For maximum value of $x$we must have $n=6$ or $n=5$

Thus $x_{max}={}^{11}C_6=\frac{11!}{5!6!}=\frac{11!}{y_{min}}$

$\Rightarrow y_{min}=5!6!$i.e. $n=6$or $n=5$

Hence the correct option is option(b) i.e. $5$